CODEWITHSUNDEEP
recursionscheme
Penrose5833
Penrose5833

Reputation: 107

Scheme Error: #t is not a function [subset?, (anon)]

I am receiving two errors and im clearly not understanding something here. The first comes when the inputs are both sets. This is the error it gives me:

Error: #t is not a function [subset?, (anon)]

The second comes when A or S is not a set and it his the else statement. I get this: 

Error: execute: unbound symbol: "else" [subset?, is-set?]

My code is as follows:

(define (is-set? L)
    (if (list? L)
        (cond
            ((null? L) #t)
            ((member (car L)(cdr L)) #f)
            (else (is-set? (cdr L)))
        )
    (else #f)
    )
)



(define (subset? A S)
    (if ((is-set? A) and (is-set? S))
        (cond
            ((null? A) #t)
            ((member (car A) S)
                 (subset? (cdr A) S))
          (else #f))
    (else #f))
)

(subset? '(1 3 5) '(1 2 3 5 6))

;;(subset? '(1 3 5) 1)

The is-set? function works and has the exact same syntax "((null? L) #t)" as "((null? A) #t)" Does it have something to do with the fact that subset? takes two parameters?

I am testing this on https://repl.it/languages/scheme im not sure if that would make a difference either.

thanks

Upvotes: 0

Views: 538

Answers (1)

sepp2k
sepp2k

Reputation: 370132

((is-set? A) and (is-set? S))

Scheme doesn't have infix operators. (exp exp*) is the syntax for a function call where the first expression evaluates to a function, which is then called with the values of the other expressions as arguments.

So the above is interpreted as a function call where (is-set? A) is the function and and and (is-set? S) are the arguments. The first problem with that is that (is-set? A) isn't actually a function (it's a Boolean), which is what the error message is telling you.

(else #f)

if does not use an else keyword. The else case of an if is simply given as the second expression to if. That is, instead of (if condition exp1 (else exp2)), you should write (if condition exp1 exp2).

The reason that you don't run into the same error with your first function is that you never call is-set? with a non-list, so it never reaches the else-case and never finds the error.

Upvotes: 1

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